# How to simplify this problem?

#### mchesnutt

##### New member
I am able to put this into wolfram and find the answer but I need to understand how this answer is found.
The problem is 1- ((tan^2 x)/(sec^2 x))
I know the solution is 1-sinx^2 could someone please explain this for me?

#### MarkFL

##### Super Moderator
Staff member
Hello, and welcome to FMH!

Consider the following:

$$\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}=\sin(x)\sec(x)$$

Can you proceed?

#### mchesnutt

##### New member
I am still pretty lost, I don't know what I am doing with the squared parts of the problem. The whole thing is really throwing me off.

#### Subhotosh Khan

##### Super Moderator
Staff member
In response #2, you were told:

tan(x) = sin(x) * sec(x)

What do you get if you "square" above?

tan2(x) = ?

What do you get if divide the above by sec2(x)?

$$\displaystyle \dfrac{\tan^2(x)}{\sec^2(x)}$$ = ?

Now evaluate:

1- $$\displaystyle \dfrac{\tan^2(x)}{\sec^2(x)}$$ = ?

#### mchesnutt

##### New member
I just went over some notes and realized I missed a section on reduction formulas. That is why none of this is making sense..
So, tan^2 x = (1-cos(2x))/(1+cos(2x)) , I know what the reduction formula is for sin, cos, and tan now, but not sec.

#### Subhotosh Khan

##### Super Moderator
Staff member
I just went over some notes and realized I missed a section on reduction formulas. That is why none of this is making sense..
So, tan^2 x = (1-cos(2x))/(1+cos(2x)) , I know what the reduction formula is for sin, cos, and tan now, but not sec.
Read response #2 and #3 and act on it

#### Jomo

##### Elite Member
In my opinion it would be easier to write (tan^2 x)/(sec^2 x) in terms of sin(x) and cos(x). Then everything will come together.